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Sonia Ayers 2022-07-09 Answered
Given
I n = 1 1 ( 1 x 2 ) n cos ( θ x ) d x ,
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Answers (1)

Tanner Hamilton
Answered 2022-07-10 Author has 12 answers
Looks like it will be easier to prove it using strong induction on n
Show the base case holds. Assume the induction hypothesis for k n, ie,
( ) k n : θ 2 k + 1 I k = k ! ( P k ( θ ) sin ( θ ) + Q k ( θ ) cos ( θ ) )
Now, show the inductive step for n + 1, ie,
θ 2 ( n + 1 ) + 1 I n + 1 = θ 2 n + 1 θ 2 I n + 1 = ( 1 ) θ 2 n + 1 ( ( 2 n + 2 ) ( 2 n + 1 ) I n ( 2 n + 2 ) ( 2 n ) I n 1 ) = ( 2 n + 2 ) ( 2 n + 1 ) [ θ 2 n + 1 I n ] ( 2 n + 2 ) ( 2 n ) θ 2 [ θ 2 ( n 1 ) + 1 I n 1 ] = ( ) ( 2 n + 2 ) ( 2 n + 1 ) n ! ( P n ( θ ) sin θ + Q n ( θ ) cos θ ) θ 2 ( 2 n + 2 ) ( 2 n ) ( ( n 1 ) ! P n 1 ( θ ) sin θ + Q n 1 cos θ ) = 2 ( 2 n + 1 ) ( n + 1 ) ! ( P n ( θ ) sin θ + Q n ( θ ) cos θ ) 4 ( n + 1 ) ! ( θ 2 P n 1 ( θ ) sin θ + θ 2 Q n 1 cos θ ) = ( n + 1 ) ! ( P n + 1 ( θ ) sin θ + Q k + 1 ( θ ) cos θ )
where P n + 1 ( θ ) := 2 ( 2 n + 1 ) P n ( θ ) 4 θ 2 P n 1 ( θ ) and Q n + 1 ( θ ) := 2 ( 2 n + 1 ) Q n ( θ ) 4 θ 2 Q n 1 ( θ ) and we can finish noting that P n + 1 and Q n + 1 are both polynomials with integer coefficients and degree at most 2 n + 2 = 2 ( n + 1 )
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